# What is the definable functor associated to an algebraic scheme (model theory of valued fields)

I have a very basic question regarding algebraic model theory. I am trying to read Espaces de Berkovich, polytopes, squelettes et théorie des modèles (MSN) by Antoine Ducros. The relevant section is Section 0.31. Let me give the setup: Let $$K$$ be a valued field and let $$\mathscr{C}_K$$ be the category of non-trivially valued, algebraically closed field extensions of $$K$$. We write $$\mathscr{L}_\mathrm{val}$$ for the language of valued fields, having three sorts (the valued field, its residue field and the value group).

Let $$\mathscr{S}$$ be a product of sorts. For each model of the theory of non-trivially valued, algebraically closed field extensions of $$K$$, i.e. for each $$F \in \mathscr{C}_K$$, we can then make sense of $$\mathscr{S}(F)$$, and this gives a functor $$\mathscr{S} \colon \mathscr{C}_K \to \mathrm{Set}$$. If $$\Phi(x)$$ is a formula of $$\mathscr{L}_\mathrm{val}$$ (with parameters in $$K$$), whose free variables $$x$$ live in $$\mathscr{S}$$, then the subfunctor of $$\mathscr{S}$$ mapping $$F$$ to $$\{x \in \mathscr{S}(F) \mid \Phi(x)\}$$ is called a definable subfunctor of $$\mathscr{S}$$.

In general, an abstract functor from $$\mathscr{C}_K$$ to $$\mathrm{Set}$$ is called definable if it is isomorphic to a definable subfunctor of some product of sorts.

Now it is claimed that a scheme of finite type $$\mathscr{X}$$ over $$K$$ gives rise to a definable functor on $$\mathscr{C}_K$$. I don't understand why this is true. If $$\mathscr{X}$$ is affine, then I believe that the associated functor should simply be the “functor of points”. Namely, if $$\mathscr{X}$$ is isomorphic to $$\mathrm{Spec}(K[T_1, \dots, T_n] / \langle f_1, \dots, f_r\rangle)$$, then its associated functor is isomorphic to the definable subfunctor of $$F \mapsto F^n$$, given by the equations $$f_i$$. I have no idea, what is meant in the non-affine case. I would be very glad about some clarification. Also, if some of my writing above hints at some misunderstanding on my side, please let me know.

• If I understand “gluing affine charts” correctly, then you are saying that also for general $\mathscr{X}$, the associated functor is the functor of points, sending $F$ to $\mathscr{X}(F) = \mathrm{Hom}_K(\mathrm{Spec}(F), \mathscr{X})$, right? Could you maybe indicate for some simple non-affine scheme like $\mathbf{P}^1_K$, how the bijection that you get to a definable functor would look like? – Jakob Werner Jun 6 '20 at 18:22
• Indeed, the functor is isomorphic to the functor of points. In the case of $\mathbf{P}_K^n$, the functor of points is isomorphic to the functor sending $F$ to the set of $n$-dimensional subspaces of $F^{n+1}$. The usual equivalence relation on $F^{n+1}$ given by $(x_0,\ldots, x_n)\sim(y_0,\ldots,y_n)$ if and only if there is $\lambda\in F^*$ such that $\lambda x_i=y_i$ for each $i\in \{0,\ldots,n\}$ is the corresponding definable equivalence relation. – Cubikova Jun 8 '20 at 9:47